LightOJ 1038 Race to 1 Again expectation of memorizing dp

Source: Internet
Author: User

LightOJ 1038 Race to 1 Again expectation of memorizing dp

Question link: Click the open link

1038-Race to 1 Again
PDF (English) Statistics Forum
Time Limit: 2 second (s) Memory Limit: 32 MB

Rimi learned a new thing about integers, which is-any positive integer greater1Can be divided by its divisors. So, he is now playing with this property. He selects a numberN. And he callthisD.

In each turn he randomly chooses a divisorD (1 to D). Then he dividesDBy the number to obtain newD. He repeats this procedureDBecomes1. What is the expected number of moves requiredNTo become1.


Input starts with an integerT (≤ 10000), Denoting the number of test cases.

Each case begins with an integerN (1 ≤ N ≤ 105).


For each case of input you have to print the case number and the expected value. Errors less10-6Will be ignored.

Sample Input Output for Sample Input





Case 1: 0

Case 2: 2.00

Case 3: 3.0333333333

After the equation is written, it is found that dp [n] = *** + dp [n] * C,

Moving dp [n] to one side is a common recursive formula.

    using namespace std;#define N 100005double dp[N];int main(){    dp[1] = 0;    dp[2] = 2;    for(int i = 3; i < N; i++) {        dp[i] = 0;        int tmp = 0;        for(int j = 1; j*j <= i; j++)        {            if(i % j == 0)            {                dp[i] += dp[j];                tmp++;                if(j != i/j && i/j != i)                {                    dp[i] += dp[i/j];                    tmp++;                }            }        }        dp[i] = ( dp[i] + tmp+1)/tmp;    }    int n, Cas=1, T, i; scanf("%d",&T);    while(T--)    {        scanf("%d", &n);        printf("Case %d: %.10f\n", Cas++, dp[n]);    }    return 0;}

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